At equilibrium, the concentrations in this system were found to be [n2]=[o2]=0.200 m and [no]=0.600 m. n2(g)+o2(g)↽−−⇀2no(g) if more no is added, bringing its concentration to 0.900 m, what will the final concentration of no be after equilibrium is re‑established?

0.84M

Explanation:

Hello,

At first, the equilibrium constant should be computed because the whole situation is at the same temperature so it is suitable for the new condition, thus:

Now, the new equilibrium condition, taking into account the change x, becomes:

Nevertheless, since the addition of NO implies that the equilibrium is leftward shifted, we should change the equilibrium constant the other way around:

Thus, we arrange the equation as:

Finally, the new concentration is:

Best regards.

neutrons are in the middle of the atom and the electrons, which are much smaller.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years